For the reaction, `2NO_2(g) hArr 2NO(g)+O_2(g) (K_c=1.8xx10^(-6)"at"184^@C)`
`(R=0.0831 L "atm" "mol"^(-1) K^(-1))`
When `K_P` and `K_c` are compared at `184^@C` it is found that

Options

A. Whether `K_P` is greater than, less then or equal to `K_c` depends upon the total gas pressure
B. `K_P=K_C`
C. `K_P` is less than `K_C`
D. `K_P` is greater than `K_C`

Answer

To solve the problem of comparing \( K_p \) and \( K_c \) for the reaction \( 2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g) \) at \( 184^\circ C \), we will follow these steps:

### Step 1: Write the relationship between \( K_p \) and \( K_c \)

The relationship between \( K_p \) and \( K_c \) is given by the formula:
\[ K_p = K_c \cdot R T^{\Delta n_g} \]
where:
- \( R \) is the ideal gas constant (0.0831 L atm mol\(^{-1}\) K\(^{-1}\)),
- \( T \) is the temperature in Kelvin,
- \( \Delta n_g \) is the change in the number of moles of gas (number of moles of gaseous products minus number of moles of gaseous reactants).

### Step 2: Calculate \( \Delta n_g \)

For the reaction:
- Products: \( 2NO(g) + O_2(g) \) gives a total of 3 moles of gas.
- Reactants: \( 2NO_2(g) \) gives a total of 2 moles of gas.

Thus, we can calculate \( \Delta n_g \):
\[ \Delta n_g = \text{(number of moles of products)} - \text{(number of moles of reactants)} = 3 - 2 = 1 \]

### Step 3: Convert the temperature to Kelvin

To use the formula, we need to convert the temperature from Celsius to Kelvin:
\[ T = 184 + 273 = 457 \, K \]

### Step 4: Substitute values into the equation

Now we can substitute the values into the equation for \( K_p \):
\[ K_p = K_c \cdot R T^{\Delta n_g} \]
Substituting the known values:
\[ K_p = K_c \cdot (0.0831) \cdot (457)^1 \]

### Step 5: Calculate \( K_p \)

First, calculate \( R \cdot T \):
\[ R \cdot T = 0.0831 \cdot 457 \approx 38.03 \]
Now substituting \( K_c = 1.8 \times 10^{-6} \):
\[ K_p = (1.8 \times 10^{-6}) \cdot 38.03 \approx 6.84 \times 10^{-5} \]

### Step 6: Compare \( K_p \) and \( K_c \)

Since \( K_p \approx 6.84 \times 10^{-5} \) and \( K_c = 1.8 \times 10^{-6} \), we can see that:
\[ K_p > K_c \]

### Conclusion

Thus, we conclude that \( K_p \) is greater than \( K_c \) at \( 184^\circ C \).

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